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I was taking a calculus competitive test, and I encountered this problem:

Henrik is randomly choosing numbers between $0$ and $1$ until the sum of all of the numbers that he has chosen exceeds $1$. What is the expected number of numbers Henrik will choose?

I spent a good amount of time on this, but I can't even think of a way to approach the problem.

I'm currently taking AP Calculus BC, so I know a good amount of calculus, but I feel like I'm missing the intuition to tackle problems like this where calculus is applied to probability.

Any help would be appreciated.

Bhaskar
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  • Are the numbers assumed to be uniformly distributed over $(0,1)$? – user Apr 20 '18 at 17:39
  • Yes, the numbers are assumed to be uniformly distributed over $(0,1)$. – Bhaskar Apr 20 '18 at 17:40
  • Have you tried the brute force method? Computing the expectation as $E(T)=\sum_{k\geq 0} kP(T=k)$? Or $E(T)=\sum_{k\geq 0} P(T\geq k)$? For what it is worth the sum of uniform follow an Irwin-Hall distribution – Frostic Apr 20 '18 at 17:41
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    I made a program to get the answer, and it yielded approximately $e$, but I want to know how to achieve this result mathematically. – Bhaskar Apr 20 '18 at 17:48
  • Nice, $e$ makes me think of characteristic functions. I will look if this can lead to something. – Frostic Apr 20 '18 at 17:49
  • The solutions at the link are way too complicated. – Denote by $E(x)$ the expected number of additional drawings when you are short by $x>0$. If $x\leq1$ then $$E(x)=1+\int_0^1 E(x−t)dt=1+\int_0^x E(\tau)>d\tau\ .$$ This implies $E'(x)=E(x)$ $(0<x\leq1)$, and together with $\lim_{x\to0+} E(x)=1$ we obtain $E(x)=e^x$, hence $E(1)=e$. – Christian Blatter Apr 21 '18 at 07:50

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I'm not sure how the question was intended to be answered but I suspect they wanted you to a formula for the expected sum.

From probability it would be the sum of the expected numbers randomly drawn so n (the number of draws) times the expected (average) number seen on a draw. This is where calculus comes in. If you are randomly drawing numbers from 0 to 1, to find the expected (average) number you draw, you integrate x times probability of drawing x over the range from 0 to 1. With all probabilities equally likely because it is a random draw, this means you integrate x times 1 dx from 0 to 1 to find the average number drawn. Then, take that answer and determine what n is so that n times that average number drawn is greater than 1.

Maggie Myers
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